Add last missing content for KW23
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LinAlg2.tex
256
LinAlg2.tex
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@ -3936,26 +3936,128 @@ Wir haben eine echte Verallgemeinerung.
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A A^\dagger A = U^T \Sigma \underbrace{V V^T}_I \Sigma^\dagger \underbrace{U U^T}_I \Sigma V
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= U^T \underbrace{\Sigma \Sigma^\dagger \Sigma}_\Sigma V = U^T \Sigma V = A
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\]
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\item[$\impliedby$:] Steht noch aus
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% \begin{itemize}
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% \item \begin{equation} \label{eq:3.6.3.1}
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% \begin{aligned}
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% \ker(\alpha) = \ker(\alpha^\dagger \circ \alpha) && \im(\alpha) = \im(\alpha \circ \alpha^\dagger) \\
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% \ker(\alpha^\dagger) = \ker(\alpha \circ \alpha^\dagger) && \im(\alpha^\dagger) = \im(\alpha^\dagger \circ \alpha)
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% \end{aligned}
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% \end{equation}
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% \tl UE\br\,:
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% $\ker(\alpha) \subseteq \ker(\alpha^\dagger \circ \alpha) \subseteq \ker(\alpha \circ \alpha^\dagger
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% \circ \alpha) = \ker(\alpha) \implies \ker(\alpha) = \ker(\alpha \circ \alpha^\dagger \circ \alpha)$
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% \item $\nu := \alpha^\dagger \circ \alpha$ ist Orthogonalprojektion auf $\ker(\alpha)^\bot$
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% \item $\nu$ selbstadjungiert $\implies \ker(v) \bot \im(v)$
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% \item $\nu \circ \nu = \alpha^\dagger \circ \underbrace{\alpha \circ \alpha^\dagger \circ \alpha}_\alpha
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% = \alpha^\dagger \circ \alpha = \nu$
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% \item $\forall u \in \im(\nu), v \in V$: \[
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% \inner{\nu(v) - v}u = \inner{\nu(v) - v}{\nu(w)} =
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% \inner{\nu^2(v) - \nu(v)}{w} = \inner{0}{w} = 0
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% \]
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% \end{itemize}
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\item[$\impliedby$:]
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\begin{itemize}
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\item Sei $\alpha \in \Hom(V, W), \alpha^\dagger \in \Hom(W, V)$
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\begin{equation} \label{eq:3.6.3.1}
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\begin{aligned}
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\ker(\alpha) = \ker(\alpha^\dagger \circ \alpha) & &
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\im(\alpha) = \im(\alpha \circ \alpha^\dagger) \\
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\ker(\alpha^\dagger) = \ker(\alpha \circ \alpha^\dagger) & &
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\im(\alpha^\dagger) = \im(\alpha^\dagger \circ \alpha)
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\end{aligned}
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\end{equation}
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\tl UE\br\,:
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$\ker(\alpha) \subseteq \ker(\alpha^\dagger \circ \alpha) \subseteq \ker(\alpha \circ \alpha^\dagger
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\circ \alpha) = \ker(\alpha) \implies \ker(\alpha) = \ker(\alpha \circ \alpha^\dagger \circ \alpha)$
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\item $\nu := \alpha^\dagger \circ \alpha$ erfüllt $\nu \circ \nu$ und ist selbstadjungiert
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? für $\nu' := \alpha \circ \alpha^\dagger$ \\
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$\implies \underbrace{\ker(\nu)}_{=\ker(\alpha)} \bot \im(\nu)$
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[Sei $v\in \ker(\nu), w = \nu(v) \in \im(\nu) \implies \inner vw = \inner{\nu(v)}{w} = 0$] \\
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$\implies$
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\begin{enumerate}[label=\alph*)]
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\item $\nu(v) \in \im(\nu)$
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\item $\forall u \in \im(\nu), v \in V:$
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\begin{align*}
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\inner{\nu(v) - v}{u} & = \inner{\nu(v) - v}{\nu(w)} = \inner{\nu^2(v) - \nu(v)}{w} \\
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& = \inner{\nu(v) - \nu(v)}{w} = 0
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\end{align*}
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\end{enumerate}
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$\implies (b_1, \dots, b_n)$ ONB mit $\langle b_{r+1}, \dots, b_n \rangle_V = \ker(\nu) =
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\ker(\alpha)$
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\begin{align*}
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& \sum_{i=1}^n \lambda_i b_i & & \overset{\nu}{\mapsto} \sum_{i=1}^r \lambda_i b_i
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\text{[$\nu$ ist orthogonale Projektion auf $\ker(\alpha)^\bot$]} \\
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\text{Analog:} & \sum_{i=1}^m \mu_i b_i' & & \overset{\nu'}{\mapsto} \sum_{i=1}^r \mu_i b_i'
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\text{[$\nu'$ ist orthogonale Projektion auf $\im(\alpha)$]} \\
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\end{align*}
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\[
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\implies {}_B M(\alpha^\dagger \circ \alpha)_B =
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\left(\begin{smallmatrix}
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1 \\
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& \ddots \\
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& & 1 \\
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& & & 0 \\
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& & & & \ddots \\
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& & & & & 0
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\end{smallmatrix}\right),\; {}_{B'} M(\alpha \circ \alpha^\dagger)_{B'} =
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\left(\begin{smallmatrix}
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1 \\
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& \ddots \\
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& & 1 \\
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& & & 0 \\
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& & & & \ddots \\
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& & & & & 0
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\end{smallmatrix}\right)
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\]
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\begin{equation}
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\label{eq:3.6.3.2}
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\begin{split}
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\implies
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\underbrace{\left(\begin{smallmatrix}
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1 \\
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& \ddots \\
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& & 1 \\
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& & & 0 \\
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& & & & \ddots \\
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& & & & & 0
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\end{smallmatrix}\right)}_n &= {}_B M(\alpha^\dagger)_{B'} \cdot {}_{B'} M(\alpha)_B = \\
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&=
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\begin{pmatrix}
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a_{11}^\dagger & \dots & a_{1m}^\dagger \\
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\vdots & & \vdots \\
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a_{n1}^\dagger & \dots & a_{nm}^\dagger
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\end{pmatrix}
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\underbrace{\left.\left(\begin{smallmatrix}
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s_1 \\
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& \ddots \\
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& & s_r \\
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& & & 0 \\
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& & & & \ddots \\
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& & & & & 0
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\end{smallmatrix}\right)\right\}}_n \scriptstyle{m}
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\end{split}
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\end{equation}
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\begin{align*}
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\underbrace{
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\left(\begin{smallmatrix}
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1 \\
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& \ddots \\
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& & 1 \\
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& & & 0 \\
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& & & & \ddots \\
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& & & & & 0
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\end{smallmatrix}\right)
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}_m & = {}_{B'} M(\alpha)_B \cdot {}_B M(\alpha^\dagger)_{B'} = \\
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& =\scriptstyle{m}\underbrace{\left\{\left(\begin{smallmatrix}
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s_1 \\
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& \ddots \\
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& & s_r \\
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& & & 0 \\
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& & & & \ddots \\
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& & & & & 0
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\end{smallmatrix}\right)\right.}_n
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\begin{pmatrix}
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a_{11}^\dagger & \dots & a_{1m}^\dagger \\
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\vdots & & \vdots \\
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a_{n1}^\dagger & \dots & a_{nm}^\dagger
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\end{pmatrix}
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\end{align*}
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Es gilt $\ker(\alpha^\dagger) = \ker(\nu') = \im(\alpha)^\bot
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= \langle b_{r+1}', \dots, b_r' \rangle \implies a^\dagger_{i\_} = 0 \forall i > r$
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\[
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\implies {}_B M(\alpha^\dagger)_{B'} =
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\begin{pmatrix}
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\begin{smallmatrix}
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a_{11}^\dagger & \dots & a_{1r}^\dagger \\
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\vdots & & \vdots \\
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a_{r1}^\dagger & \dots & a_{rr}^\dagger
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\end{smallmatrix} & 0 \\
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0 & 0
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\end{pmatrix}
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\text{ und } a_{ij}^\dagger = \delta_{ij} \frac{1}{s_i} \text{ wegen \ref{eq:3.6.3.2}}
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\]
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\end{itemize}
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\end{itemize}
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\end{proof}
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@ -3999,7 +4101,9 @@ $w = \sum_{i=1}^n \mu_i b_i'$
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\norm{\alpha(v) - w}^2 = \norm{\sum_{i=1}^r s_i \lambda_i b_i' - \sum_{i=1}^m \mu_i b_i'}^2 =
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\sum_{i=1}^r (s_i \lambda_i - \mu_i)^2 + \sum_{i = r+1}^m \mu_i^2
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\]
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Wird minimal wenn $\lambda_i = \frac{\mu_i}{s_i}, i \in [r]$, insbesondere für $v = \alpha^\dagger(w)$
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Wird minimal wenn $\lambda_i = \frac{\mu_i}{s_i}, i \in [r]$, das heißt das optimale $v$ ist durch
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$v^\dagger = \alpha^\dagger(w)$ gegeben.
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$\left[\text{Alle optimalen durch }v^\dagger + \sum_{j=r+1}^m \mu_j^2 = L(\alpha^* \alpha, \alpha^*(w))\right]$
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\begin{satz}
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Sei $\alpha \in \homk(V, W), \K \in \{\R,\C\}, V, W$ endlich dimensional.
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@ -4010,6 +4114,16 @@ Wird minimal wenn $\lambda_i = \frac{\mu_i}{s_i}, i \in [r]$, insbesondere für
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Alle Vektoren mit dieser Eigenschaft erfüllen die\\ \underline{Normalgleichungen}
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$\ontop{\alpha^* \alpha(v) = \alpha^*(w)}{A^* A x = A^* b}$
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\end{satz}
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\begin{proof}
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Angenommen $L(A, b) \neq \emptyset$. $\to$ Suche $v \in L(A, b)$ mit minimaler Norm:
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Sei $w \in \im(\alpha) \implies w = \sum_{j=1}^r \mu_j b_j'$
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\[
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\implies L(\alpha, w) = \left\{ \sum_{j=1}^r \frac{\mu_j}{s_j} b_j + \sum_{j=r+1}^n \lambda_j b_j:
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\lambda_{r+1}, \dots, \lambda_n \in \K\right\}
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\]
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Minimale Norm, wenn $\lambda_{r+1}, \dots, \lambda_n = 0$ das heißt für
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$v = \sum_{i=1}^r \frac{\mu_i}{s_i} b_i = \alpha^\dagger(w)$
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\end{proof}
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\begin{satz}
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Sei $\alpha \in \Hom(V, W), w \in \im(\alpha)$.
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@ -4019,47 +4133,65 @@ Wird minimal wenn $\lambda_i = \frac{\mu_i}{s_i}, i \in [r]$, insbesondere für
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\]
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\end{satz}
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%\subsubsection{Beispiel (lineare Regression)}
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%\begin{tikzpicture}
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%\end{tikzpicture}
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%$(t_i, y_i)_{i=1}^m$.
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%Suche $f: f(t_i) \sim y_i, \forall i \in [m]$
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%$f(t) = a_0 + a_1 t + a_2 t^2$
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%\[
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% \text{minimiere }
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% \sum_{i=1}^m (f(t_i) - y_i)^2 = \sum_{i=1}^m (a_0 + a_1 t_i + a_2 t_i^2 - y_i)^2 = \norm{A x -b}^2_{\K^m}
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%\]
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\subsubsection{Beispiel (lineare Regression)}
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$(t_i, y_i)_{i=1}^m$ gegeben. Wollen Polynome finden, die gut auf die Messungen passen.
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Suchen also $p: p(t_i) \sim y_i, \forall i \in [m]$ \\
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\begin{tikzpicture}[declare function={approx(\x) = 0.17 * \x * \x + -1.1 * \x + 4.1;}]
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\draw [red, thick] plot [domain=0:10,samples=144,smooth] (\x, {approx(\x)}) node[right,color=black]
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{$p(t) = a_0 + a_1 t + a_2 t^2$};
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\foreach[count=\i] \diff in {0.93,-0.56,-2.12,1.35,-0.83,0.87,-0.04,-1.16,-0.02,1.25,0.13,-0.62,2.71,-1.84,-0.24,1.64,-0.06,-0.52,-0.21,-1.24}
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\filldraw[ForestGreen] (\i * .5, {approx(\i * .5) + \diff * .6}) circle (.5mm);
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\draw[->] (0, 0) -- (10, 0);
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\draw[->] (0, 0) -- (0, 10);
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\end{tikzpicture}
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\[
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\text{minimiere }
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\sum_{i=1}^m (f(t_i) - y_i)^2 = \sum_{i=1}^m (a_0 + a_1 t_i + a_2 t_i^2 - y_i)^2 = \norm{A x -b}^2_{\K^m}
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\text{ wobei}
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\]
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\[
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A = \begin{pmatrix} 1 & t_1 & t_1^2 \\ \vdots & \vdots & \vdots \\ 1 & t_m & t_m^2 \end{pmatrix},
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x = \begin{pmatrix} a_0 \\ a_1 \\ a_2 \end{pmatrix},
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b = \begin{pmatrix} y_1 \\ \vdots \\ y_m \end{pmatrix}
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\]
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%\subsubsection{Anwendung: Ausgleichsquadrik}
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%Problem: homogenes LGS $Ax=0$. Finde $x$ mit $\norm x = 1$ und $\norm{Ax}$ minimal. \\
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%$b_1, \dots, b_n$ ONB aus EVen von $A^* A$ mit nichtnegativen EWen.
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%\begin{align*}
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% X = \sum \lambda_i b_i \implies \norm{Ax}^2 & = \inner{Ax}{Ax} \\
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% & = \inner{A^* A x}{x} =
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% \inner{\sum s_i \lambda_i b_i}{\sum \lambda_j b_j} = \sum_{i=1}^n s_i \abs{\lambda_i}^2
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%\end{align*}
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%$s_1 \le s_2 \le \dots \le s_n, \norm x = \sum \abs{\lambda_i}^2$
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%\[
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% \frac{\norm{Ax}}{\norm x} = \frac{\sum s_i \abs{\lambda_i}^2}{\sum \abs{\lambda_i}^2} \ge
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% \frac{s_1 \sum \abs{\lambda_i}^2}{\sum \abs{\lambda_i}^2} s_1
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%\]
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%$\norm x = 1 \implies \norm{Ax} \ge s_1$
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%$\norm{b_i} \implies \lambda_1, \lambda_2 = \dots = \lambda_n = 0 \implies \norm{Ab_1} = s_1 \implies b_1$
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%löst unser Minimierungsproblem. \\
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%$Q = \{(x,y) \in \R^2: \psi(x, y) = 0\}, \psi(x, y):= a_1 x^2 + a_2 xy + a_3 y^2 a_4 x + a_5 y + a_6$
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%Gegeben: $(x_i,y_i)^m_{i=1}$ Suche $x = (a_1, \dots, a_6)^T$ mit $\norm x = 1$ sodass
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%\[
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% \sum_{i=1}^m (a_1 x_i^2 + a_2 x_i y_i + a_3 y_i^2 + a_4 x_i + a_5 y_y + a_6)^2
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%\]
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%minimal.
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%$=\norm{Ax}^2, A = \begin{pmatrix}\end{pmatrix}$
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%
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%\begin{satz*}
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% Sei $A \in \K^{m \times n}$ und $b \in \K^n$ Eigenvektor von $A^* A$ zum kleinsten Eigenwert $r_1$.
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% Dann gilt
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% \[
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% \frac{\norm{Ab}}{\norm b} = \min\left\{\frac{\norm{Ax}}{\norm x}: x\in\R^n\right\} = \sqrt{r_1}
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% \]
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%\end{satz*}
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\subsubsection{Anwendung: Ausgleichsquadrik}
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Problem: homogenes LGS $Ax=0$. Finde $x$ mit $\norm x = 1$ und $\norm{Ax}$ minimal. \\
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$b_1, \dots, b_n$ ONB aus EVen von $A^* A$ mit nichtnegativen EWen.
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\begin{align*}
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X = \sum \lambda_i b_i \implies \norm{Ax}^2 & = \inner{Ax}{Ax} \\
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& = \inner{A^* A x}{x} =
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\inner{\sum s_i \lambda_i b_i}{\sum \lambda_j b_j} = \sum_{i=1}^n s_i \abs{\lambda_i}^2
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\end{align*}
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$s_1 \le s_2 \le \dots \le s_n, \norm x = \sum \abs{\lambda_i}^2$
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\[
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\frac{\norm{Ax}}{\norm x} = \frac{\sum s_i \abs{\lambda_i}^2}{\sum \abs{\lambda_i}^2} \ge
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\frac{s_1 \sum \abs{\lambda_i}^2}{\sum \abs{\lambda_i}^2} s_1
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\]
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$\norm x = 1 \implies \norm{Ax} \ge s_1$
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$\norm{b_i} \implies \lambda_1, \lambda_2 = \dots = \lambda_n = 0 \implies \norm{Ab_1} = s_1 \implies b_1$
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löst unser Minimierungsproblem. \\
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$Q = \{(x,y) \in \R^2: \psi(x, y) = 0\}$ \\
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$\psi(x, y):= a_1 x^2 + a_2 xy + a_3 y^2 a_4 x + a_5 y + a_6$ \\
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Gegeben: $(x_i,y_i)^m_{i=1}$ Suche $x = (a_1, \dots, a_6)^T$ mit $\norm x = 1$ sodass
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\[
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\sum_{i=1}^m (a_1 x_i^2 + a_2 x_i y_i + a_3 y_i^2 + a_4 x_i + a_5 y_i + a_6)^2=\norm{Ax}^2
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\]
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minimal.
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$A = \begin{pmatrix}
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x_1^2 & x_1 y_1 & y_1^2 & x_1 & y_1 & 1 \\
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\vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\
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x_m^2 & x_m y_m & y_m^2 & x_m & y_m & 1 \\
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\end{pmatrix}$ \\
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Suche $x \in \R^6$ mit $\norm x = 1$ und $\norm{Ax}$ minimal.
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$\implies x$ ist Eigenvektor von $A^* A$ zum kleinsten Eigenwert.
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\begin{satz*}
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Sei $A \in \K^{m \times n}$ und $b \in \K^n$ Eigenvektor von $A^* A$ zum kleinsten Eigenwert $r_1$.
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Dann gilt
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\[
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\frac{\norm{Ab}}{\norm b} = \min\left\{\frac{\norm{Ax}}{\norm x}: x\in\R^n\right\} = \sqrt{r_1}
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\]
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\end{satz*}
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\end{document}
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